SOLUTION: An object that is falling or vertically projected into the air has its height, in feet, above the ground given by s=-16t^2+vot=so Where s is the height, in fee, vo is the original

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: An object that is falling or vertically projected into the air has its height, in feet, above the ground given by s=-16t^2+vot=so Where s is the height, in fee, vo is the original       Log On


   

Question 88736: An object that is falling or vertically projected into the air has its height, in feet, above the ground given by s=-16t^2+vot=so Where s is the height, in fee, vo is the original velocity of the object in feet per second, t is the time the object is in motion, is seconds, and so is the height in feet, from which the object is dropped or projected. The figure shows that a ball is thrown straight up from a rooftop at an original velocity of 80 feet per second from a height of 96 feet. The ball misses the rooftop on its way down and eventually strikes the ground. Use the formula and the infromation to solve problem 89-91 I am solving even #90
s=-16t^2+80t+96
t=4
s= -16(4^2)+80(4)+96
s= -16(16)+320+96
s=-256+320+96
s= 160ft


Ruchi: Correct solution, Darlene.
The equation for this problem is s=-16t^2+80t+96. Is there any way you can figure out how many seconds the ball will take to reach the ground?
I can't quite figure out how to set up this last question and solve it.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If we set the equation equal to zero, we can find out when the ball will hit the ground
-16t%5E2%2B80t%2B96=0

So let's use the quadratic formula to solve for t:


Starting with the general quadratic

at%5E2%2Bbt%2Bc=0

the general solution using the quadratic equation is:

t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve -16%2At%5E2%2B80%2At%2B96=0 ( notice a=-16, b=80, and c=96)

t+=+%28-80+%2B-+sqrt%28+%2880%29%5E2-4%2A-16%2A96+%29%29%2F%282%2A-16%29 Plug in a=-16, b=80, and c=96



t+=+%28-80+%2B-+sqrt%28+6400-4%2A-16%2A96+%29%29%2F%282%2A-16%29 Square 80 to get 6400



t+=+%28-80+%2B-+sqrt%28+6400%2B6144+%29%29%2F%282%2A-16%29 Multiply -4%2A96%2A-16 to get 6144



t+=+%28-80+%2B-+sqrt%28+12544+%29%29%2F%282%2A-16%29 Combine like terms in the radicand (everything under the square root)



t+=+%28-80+%2B-+112%29%2F%282%2A-16%29 Simplify the square root



t+=+%28-80+%2B-+112%29%2F-32 Multiply 2 and -16 to get -32

So now the expression breaks down into two parts

t+=+%28-80+%2B+112%29%2F-32 or t+=+%28-80+-+112%29%2F-32

Lets look at the first part:

x=%28-80+%2B+112%29%2F-32

t=32%2F-32 Add the terms in the numerator
t=-1 Divide

So one answer is
t=-1



Now lets look at the second part:

x=%28-80+-+112%29%2F-32

t=-192%2F-32 Subtract the terms in the numerator
t=6 Divide

So another answer is
t=6

So our solutions are:
t=-1 or t=6

Notice when we graph -16%2Ax%5E2%2B80%2Ax%2B96 (just replace t with x), we get:

+graph%28+500%2C+500%2C+-11%2C+16%2C+-11%2C+16%2C-16%2Ax%5E2%2B80%2Ax%2B96%29+

and we can see that the roots are t=-1 and t=6. This verifies our answer

Since a negative time doesn't make sense, so our only solution is t=6. So it will take 6 seconds for the ball to hit the ground.